Optimisation of a cylindrical container

[pavadrin]

Hey
I am given the set volume for a cylindrical container and separating pricing for the material used to construct the container, the base and the side wall. The ends of the container cost $0.05 per cm2 and the side walls $0.04 cm2, and the volume is 400 mL. The question asks: What is the best design for the container i.e., measurements to ensure that cost is at a minimum.

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my working:
V = 400 mL
Ends costs $0.05 per cm2
Side wall costs $0.04 cm2
Area of ends = $$2 \pi r^2$$
Area of side wall $$2 \pi r h$$
V = 400, therefore since V = $$\pi r^2 h$$, $$h = \frac{400}{\pi r^2}$$ (1)
Cost for one container = $$0.05 (2 \pi r^2) + 0.04 (2 \pi r h)$$ (2)
Therefore substituting (1) into (2), cost = $$0.05 (2 \pi r^2) + 0.04 (2 \pi r \frac{400}{\pi r^2})$$
Differentiating the cost function gives: $$\frac {dC}{dr} = 0.2 \pi r + \frac {96}{r^2}$$
The value of $$r$$ where there is no rate of change = -5.34601847

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I am not sure if what I have done after and including the differentiation is correct. Since I have got a negative answer for the value of r, I am thinking that what I have done is wrong, as you cannot have a negative radius. Would the second derivative help with the sloving this equation, if so how?
Many thanks,
Pavadrin

 

[HallsofIvy]

Hey
I am given the set volume for a cylindrical container and separating pricing for the material used to construct the container, the base and the side wall. The ends of the container cost $0.05 per cm2 and the side walls $0.04 cm2, and the volume is 400 mL. The question asks: What is the best design for the container i.e., measurements to ensure that cost is at a minimum.

_________________________________

my working:
V = 400 mL
Ends costs $0.05 per cm2
Side wall costs $0.04 cm2
Area of ends = $$2 \pi r^2$$
Area of side wall $$2 \pi r h$$
V = 400, therefore since V = $$\pi r^2 h$$, $$h = \frac{400}{\pi r^2}$$ (1)
Cost for one container = $$0.05 (2 \pi r^2) + 0.04 (2 \pi r h)$$ (2)
Therefore substituting (1) into (2), cost = $$0.05 (2 \pi r^2) + 0.04 (2 \pi r \frac{400}{\pi r^2})$$

Go ahead and write this out: cost= [tex]0.1 \pi r^2+ \frac{32}{r}[/itex]<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Differentiating the cost function gives: $$\frac {dC}{dr} = 0.2 \pi r + \frac {96}{r^2}$$ </div> </div> </blockquote> Not quite. It gives $$\frac{dC}{dr}= 0.2 \pi r- \frac{64}{r^2}$$.<br /> Notice the negative sign! The derivative of r^-1 is -r^-2.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The value of $$r$$ where there is no rate of change = -5.34601847<br /> <br /> _________________________________<br /> <br /> I am not sure if what I have done after and including the differentiation is correct. Since I have got a negative answer for the value of r, I am thinking that what I have done is wrong, as you cannot have a negative radius. Would the second derivative help with the sloving this equation, if so how?<br /> Many thanks,<br /> Pavadrin </div> </div> </blockquote> <br /> You should have $$r^3= \frac{0.2 \pi}{32}$$.[/tex]

 

[pavadrin]

oh i sort of get it now, well at least more than before, thanks

 

[pavadrin]

thanks for the reply, however I am not sure how u got the derivative of the cost being equal to $$\frac{dC}{dr}= 0.2 \pi r- \frac{64}{r^2}$$. I am not sure on how to achieve the value of 64 as i get a value of 32. My derrivitive: $$\frac{dC}{dr}= 0.2 \pi r- \frac{32}{r^2}$$. Could u please explain how this is achieved? help greatly appriciated
Pavadrin

 

[arunbg]

Yes, I think that was a typo from Halls .

 

[HallsofIvy]

No, just stupidity!

 

[pavadrin]

okay, fair enough, we all make mistakes sooner or later
thanks anyhow